Q1 / 50 · Manipal (MET) 2022 · Chemistry
Arrange the following in increasing order of the volume (in L) occupied by them at STP
(i) 1.5 moles of $\mathrm{CO}_2$
(ii) 14 g of $\mathrm{N}_2$
(iii) $10^{21}$ molecules of oxygen
- Correct answerA(iii) $ \lt $ (ii) $ \lt $ (i)
- B(i) $ \lt $ (ii) $ \lt $ (iii)
- C(ii) $ \lt $ (i) $ \lt $ (iii)
- D(i) $ \lt $ (iii) $ \lt $ (ii)
Solution
(i) Volume occupied by 1 mole of $\mathrm{CO}_2=22.4 \mathrm{~L}$
Volume occupied by 1.5 moles of $\mathrm{CO}_2$ $\begin{aligned} & =1.5 \times 22.4 \\ & =33.6 \mathrm{~L} \end{aligned}$ (ii) Given, mass of nitrogen $(m)=14 \mathrm{~g}$
Molar mass of $\mathrm{N}_2(w)=28 \mathrm{~g}$ Number of moles of $\mathrm{N}_2=\frac{m}{w}=\frac{14}{28}=\frac{1}{2}$ $=0.5 \mathrm{~mole}$
Volume occupied by 1 mole of $\mathrm{N}_2$ at STP $=22.4 \mathrm{~L}$
Volume occupied by 0.5 mole of $\mathrm{N}_2$ at STP $\begin{aligned} & =0.5 \times 22.4 \\ & =11.2 \mathrm{~L} \end{aligned}$ (iii) Given, number of molecules of $\mathrm{O}_2=10^{21}$
Number of molecules of $\mathrm{O}_2$ in 1 mole $\begin{aligned} & =6.022 \times 10^{23} \text { molecules } \\ \text { Number of moles } & =\frac{10^{21}}{6.022 \times 10^{23}} \\ & =0.166 \times 10^{-2} \text { moles } ; \end{aligned}$
Volume occupied by $0.166 \times 10^{-2}$ mole of $\begin{aligned} \mathrm{O}_2 & =0.166 \times 10^{-2} \times 22.4 \mathrm{~L} \\ & =3 . \nVdash 2 \times 10^{-2} \\ & =0.0372 \mathrm{~L} \end{aligned}$
Therefore, correct increasing order of volume occupied by the given amount of molecules at STP is : (iii) < (ii) < (i).
Volume occupied by 1.5 moles of $\mathrm{CO}_2$ $\begin{aligned} & =1.5 \times 22.4 \\ & =33.6 \mathrm{~L} \end{aligned}$ (ii) Given, mass of nitrogen $(m)=14 \mathrm{~g}$
Molar mass of $\mathrm{N}_2(w)=28 \mathrm{~g}$ Number of moles of $\mathrm{N}_2=\frac{m}{w}=\frac{14}{28}=\frac{1}{2}$ $=0.5 \mathrm{~mole}$
Volume occupied by 1 mole of $\mathrm{N}_2$ at STP $=22.4 \mathrm{~L}$
Volume occupied by 0.5 mole of $\mathrm{N}_2$ at STP $\begin{aligned} & =0.5 \times 22.4 \\ & =11.2 \mathrm{~L} \end{aligned}$ (iii) Given, number of molecules of $\mathrm{O}_2=10^{21}$
Number of molecules of $\mathrm{O}_2$ in 1 mole $\begin{aligned} & =6.022 \times 10^{23} \text { molecules } \\ \text { Number of moles } & =\frac{10^{21}}{6.022 \times 10^{23}} \\ & =0.166 \times 10^{-2} \text { moles } ; \end{aligned}$
Volume occupied by $0.166 \times 10^{-2}$ mole of $\begin{aligned} \mathrm{O}_2 & =0.166 \times 10^{-2} \times 22.4 \mathrm{~L} \\ & =3 . \nVdash 2 \times 10^{-2} \\ & =0.0372 \mathrm{~L} \end{aligned}$
Therefore, correct increasing order of volume occupied by the given amount of molecules at STP is : (iii) < (ii) < (i).
Hence, the correct answer is option: 'A'.